The answer is : #int_0^1 x5^x dx ~~1.56#.
We have #h(x) = 5^x x = f'(x)g(x)#.
Here, #f'(x) = 5^x# and #g(x) = x#.
The antiderivative of #h(x)# is given by :
#f(x)g(x) - int f(x)g'(x)dx#.
Firstly, let's calculate the derivative of #5^x# by using the limit definition, it will help us to get its antiderivative :
#j'(x) = lim_(h->0)(j(x + h) -j(x))/h#
#(5^x)' = lim_(h->0) (5^(x+h) -5^(x))/h = lim_(h->0) (5^(x)*(5^h-1))/h#
#= 5^(x) lim_(h->0) (5^h-1)/h = 5^(x) lim_(h->0) (e^(hln(5))-1)/h#
#= 5^(x) lim_(h->0) ln(5)((e^(hln(5))-1)/(hln(5)))#
#= 5^(x) ln(5) lim_(h->0) (e^(hln(5))-1)/(hln(5))#
#=5^(x) ln(5) lim_(k->0) (e^(k)-1)/(k)# , where #k = hln(5)#
By the definition of #e#, #lim_(k->0) (e^(k)-1)/(k)=1#.
So #(5^x)' = 5^x ln(5)#.
Therefore, the antiderivative of #f'(x) = 5^x# is #f(x) = 5^x/ln(5)# because :
#(5^x/ln(5))' = ((5^x)')/ln(5) = (5^x ln(5))/ln(5) = 5^x#
And the derivative of #g(x) = x# is #g'(x) = 1#.
Thus, the antiderivative of #h(x)# is :
#H(x) = f(x)g(x) - int f(x)g'(x)dx = (5^x x)/ln(5) - int 5^x/ln(5)#
#= (5^x x)/ln(5) - 1/ln(5) int 5^x = (5^x x)/ln(5) - 5^x/ln^2(5)#
Now, we can calculate the integral :
#int_0^1 h(x) dx = [H(x)]_0^1 = H(1) - H(0)#
#= 5/ln(5) - 5/ln^2(5) - 0 + 1/ln^2(5) = 5/ln(5) - 4/ln^2(5)#
#= (5ln(5)-4)/ln^2(5)~~1.56#.
That's it!
